UK diy (uk.d-i-y) For the discussion of all topics related to diy (do-it-yourself) in the UK. All levels of experience and proficency are welcome to join in to ask questions or offer solutions.

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
  #1   Report Post  
Will Dean
 
Posts: n/a
Default Adiabatic short-circuit compliance on very short short-circuits


Hi,

This is an abstract question - but I'm just interested to see what I've
misunderstood here...

Imagine a domestic installation, with a PSC of 6kA at the incoming supply,
fused at 100A. Right next to the incoming supply, there's a consumer unit,
which contains one 20A MCB. The CU is connected to the incoming supplies
with very short 25mmsq tails (I've ignored their contribution to lowering
the fault level.)

This 20A MCB supplies a short (let's say 0.5m) piece of 2.5mmsq T&E, which
supplies a single socket next to the consumer unit.

My calculations suggest that the L-N fault current at the socket is going to
be about 5kA, and for adiabatic I^2t or (kS)^2 compliance, we would need to
break a short circuit in around 3ms max to avoid damaging the 2.5mm cable.

This seems to be well off the bottom of graphs for MCB response times. All
examples I can find around the place of using the adiabatic temperature rise
equations give answers which are conveniently above 0.1ms, which allows
slightly glib commentary about MCBs always tripping in under 100ms, etc.

Although this was a completely contrived example, it actually applies to any
S/C fault on any final circuit which occurs sufficiently close to the CU.

Something like this:

http://groups-beta.google.com/group/...e=source&hl=en

contains an example adiabatic compliance calculation, but it appears to me
that it only considers a short occurring at the end of the final circuit,
not close to the beginning. (It's also considering L-PE rather than L-N,
but I don't think that's an important distinction here)

I'm clear why *load* is considered only at the end of a circuit, but not why
fault conditions are being calculated for the end of the circuit.

Given that fuse and MCB time vs. current graphs don't tend to go down to
single-millisecond levels, should one actually be looking at I2t let-through
graphs and comparing that with k2S2?

TIA,

Will (hoping that Andy and Andrew aren't both on holiday!)




  #2   Report Post  
Andrew Gabriel
 
Posts: n/a
Default

In article ,
"Will Dean" writes:

Hi,

This is an abstract question - but I'm just interested to see what I've
misunderstood here...

Imagine a domestic installation, with a PSC of 6kA at the incoming supply,
fused at 100A. Right next to the incoming supply, there's a consumer unit,
which contains one 20A MCB. The CU is connected to the incoming supplies
with very short 25mmsq tails (I've ignored their contribution to lowering
the fault level.)

This 20A MCB supplies a short (let's say 0.5m) piece of 2.5mmsq T&E, which
supplies a single socket next to the consumer unit.

My calculations suggest that the L-N fault current at the socket is going to
be about 5kA, and for adiabatic I^2t or (kS)^2 compliance, we would need to
break a short circuit in around 3ms max to avoid damaging the 2.5mm cable.


Really you want Andy for this, but the 5kA sounds wrong to me.

Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms.
Add to that the 0.5m of 2.5mmsq T&E, which will have a
resistance of 0.09 ohms, and we have 0.13 ohms.
240V across 0.13 ohms is only 1.846kA.

Being a physicist rather than an electronic engineer, I'll
work out the temperature rise from first principles rather
than using the method in BS7671.

Let's assume fastest MCB trip time possible is half a cycle, 0.01 sec.
500cm of 2.5mm^2 cable during a fault current of 1.846kA for 0.01 sec.
Mass of copper is 8.96g/cm^3 * 0.025cm^2 * 500cm * 2 = 224g,
Specific heat capacity of copper = 0.385J/g/K...

Power dissipated per 500cm:

P = I^2 * R = 1846^2 * 0.09 = 306kW

Energy dissipated per 500cm during 0.1 sec:

306kW * 0.01 = 3060J.

Temp rise = 3060 / 0.385 / 224 = 35.5 Centigrade

Thus a fault current of 1.846kA for 0.1 sec will result in a temperature
rise of about 35.5 Centigrade in the copper conductors. If cable was
previously within its correct operating temperature range (i.e. = 70 C),
it will survive this without damage (i.e. not exceed 160 C for PVC).

--
Andrew Gabriel
  #3   Report Post  
Andrew Gabriel
 
Posts: n/a
Default

[cancaled earlier posting due to mistake in calculation]

In article ,
"Will Dean" writes:

Hi,

This is an abstract question - but I'm just interested to see what I've
misunderstood here...

Imagine a domestic installation, with a PSC of 6kA at the incoming supply,
fused at 100A. Right next to the incoming supply, there's a consumer unit,
which contains one 20A MCB. The CU is connected to the incoming supplies
with very short 25mmsq tails (I've ignored their contribution to lowering
the fault level.)

This 20A MCB supplies a short (let's say 0.5m) piece of 2.5mmsq T&E, which
supplies a single socket next to the consumer unit.

My calculations suggest that the L-N fault current at the socket is going to
be about 5kA, and for adiabatic I^2t or (kS)^2 compliance, we would need to
break a short circuit in around 3ms max to avoid damaging the 2.5mm cable.


Really you want Andy for this, but the 5kA sounds wrong to me.

Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms.
Add to that the 0.5m of 2.5mmsq T&E, which will have a
resistance of 0.09 ohms, and we have 0.13 ohms.
240V across 0.13 ohms is only 1.846kA.

Being a physicist rather than an electronic engineer, I'll
work out the temperature rise from first principles rather
than using the method in BS7671.

Let's assume fastest MCB trip time possible is half a cycle, 0.01 sec.
Specific heat capacity of copper = 0.385J/g/K
Density of copper = 8.96g/cm^3
So, mass of copper in 500cm of 2.5mm^2 cable is
8.96g/cm^3 * 0.025cm^2 * 500cm * 2 = 224g

Power dissipated per 500cm:

P = I^2 * R = 1846^2 * 0.09 = 306kW

Energy dissipated per 500cm during 0.01 sec:

306kW * 0.01 = 3060J.

Temp rise = 3060 / 0.385 / 224 = 35.5 Centigrade

Thus a fault current of 1.846kA for 0.01 sec will result in a temperature
rise of about 35.5 Centigrade in the copper conductors. If cable was
previously within its correct operating temperature range (i.e. = 70 C),
it will survive this without damage (i.e. not exceed 160 C for PVC).

--
Andrew Gabriel
  #5   Report Post  
Will Dean
 
Posts: n/a
Default

"Andrew Gabriel" wrote in message
.. .

Really you want Andy for this, but the 5kA sounds wrong to me.

Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms.


Yup, agree that.

Add to that the 0.5m of 2.5mmsq T&E, which will have a
resistance of 0.09 ohms,


I don't get that - 90 mOhm.
I think it's more like 7 mOhm.

Cheers,

Will




  #6   Report Post  
Andrew Gabriel
 
Posts: n/a
Default

In article ,
"Will Dean" writes:
"Andrew Gabriel" wrote in message
.. .

Really you want Andy for this, but the 5kA sounds wrong to me.

Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms.


Yup, agree that.

Add to that the 0.5m of 2.5mmsq T&E, which will have a
resistance of 0.09 ohms,


I don't get that - 90 mOhm.
I think it's more like 7 mOhm.


Yes, I got it wrong.
It should be 0.009 ohms -- Table 4D2B claims 18 mOhms/metre.
I'm not sure where your 7 mOhm comes from, although it's a
lot nearer than my 90 mOhms ;-)

There's another mistake too -- I calculated the temperature
rise for the mass of copper in 5m of cable rather than 0.5m.

These two combined make the result way off. I'll do the
calculation and post again.

--
Andrew Gabriel
  #7   Report Post  
Will Dean
 
Posts: n/a
Default

"Andrew Gabriel" wrote in message
.. .

Yes, I got it wrong.
It should be 0.009 ohms -- Table 4D2B claims 18 mOhms/metre.
I'm not sure where your 7 mOhm comes from, although it's a
lot nearer than my 90 mOhms ;-)


I suspect the 9 vs 7 is temperature related.
Table 9A in the OSG gives 14.82/2 = 7.41
Using 16.8nOhm-metres (the first figure I stumbled across on the internet -
I don't know what temperature it's for) gives something around 7-ish, too.

There's another mistake too -- I calculated the temperature
rise for the mass of copper in 5m of cable rather than 0.5m.


A physicist, you said? Hmm.

Cheers,

Will



  #8   Report Post  
Andrew Gabriel
 
Posts: n/a
Default

In article ,
(Andrew Gabriel) writes:
These two combined make the result way off. I'll do the
calculation and post again.


Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms.
Add to that the 0.5m of 2.5mmsq T&E, which will have a
resistance of 0.009 ohms, and we have 0.049 ohms.
240V across 0.049 ohms is 4.878kA.

Let's assume fastest MCB trip time possible is half a cycle, 0.01 sec.
Specific heat capacity of copper = 0.385J/g/K
Density of copper = 8.96g/cm^3
So, mass of copper in 50cm of 2.5mm^2 cable is
8.96g/cm^3 * 0.025cm^2 * 50cm * 2 = 22.4g

Power dissipated per 50cm:

P = I^2 * R = 4878^2 * 0.009 = 214kW

Energy dissipated per 50cm during 0.01 sec:

214kW * 0.01 = 2140J.

Temp rise = 2140 / 0.385 / 22.4 = 248 Centigrade

Thus a fault current of 4878A for 0.01 sec will result in a temperature
rise of about 248 Centigrade in the copper conductors, which will
damage or wreck the cable.

So let's work backwards to identify the maximum fault current 2.5mm^2
T&E can handle...

Max fault current temperature rise allowed is 90C.
Mass of copper per metre = 44.8g
Max energy per metre = 90 * 44.8 * 0.385 = 1552 Joules.
Max power per metre for a 0.01 second fault = 1552 / 0.01 = 155200 Watts.
Max current = sqrt(P/R) = sqrt(155200 / 0.018) = 2936A

So you are always going to get some length of 2.5mm^2 T&E which
is inadiquately protected if the PSC at supply end is greater
than 2936A. In the case of your 6kA PSC supply and 2.5mm^2 T&E,
that's going to be the first 2.3 metres of cable (if I did the
calculation right;-)

--
Andrew Gabriel
  #9   Report Post  
Owain
 
Posts: n/a
Default

Will Dean wrote:
"Andrew Gabriel" wrote
There's another mistake too -- I calculated the temperature
rise for the mass of copper in 5m of cable rather than 0.5m.

A physicist, you said? Hmm.


Physicist mathematician :-)

patronising:-) I think he did very well /patronising:-)

Owain

  #10   Report Post  
Will Dean
 
Posts: n/a
Default

"Andrew Gabriel" wrote in message
.. .

So you are always going to get some length of 2.5mm^2 T&E which
is inadiquately protected if the PSC at supply end is greater
than 2936A. In the case of your 6kA PSC supply and 2.5mm^2 T&E,
that's going to be the first 2.3 metres of cable (if I did the
calculation right;-)


OK, we're sort of agreeing now.

Given that all circuits of = 2.3 metres are susceptible to faults WITHIN
the first 2.3 metres, this situation must pertain for almost every final
circuit in real life.

I can only assume that one must check I2t let-through figures for the
protection device, rather than disconnection time graphs.

What I don't understand is why some of the examples I've seen seem to
include the final circuit's own loop impedance in this calculation - that
seems to be a mistake.

Cheers,

Will




  #11   Report Post  
Ed Sirett
 
Posts: n/a
Default

On Fri, 05 Aug 2005 18:06:10 +0100, Will Dean wrote:

"Andrew Gabriel" wrote in message
.. .

So you are always going to get some length of 2.5mm^2 T&E which
is inadiquately protected if the PSC at supply end is greater
than 2936A. In the case of your 6kA PSC supply and 2.5mm^2 T&E,
that's going to be the first 2.3 metres of cable (if I did the
calculation right;-)


OK, we're sort of agreeing now.

Given that all circuits of = 2.3 metres are susceptible to faults WITHIN
the first 2.3 metres, this situation must pertain for almost every final
circuit in real life.

In real life PSC of 6kA are uncommon although theoretically possible.
A plausible scenario would be a cleaners' socket in a meter cupboard on
the ground floor of a block of flats containing the sub-station in the
basement.

I can only assume that one must check I2t let-through figures for the
protection device, rather than disconnection time graphs.

What I don't understand is why some of the examples I've seen seem to
include the final circuit's own loop impedance in this calculation - that
seems to be a mistake.


I thought the resistance of the MCB might come to our rescue but I've just
measured one and it's DC resistance is less than 0.01 ohms as measured by
good test gear. Likewise the meter might help out a bit? However the AC
impedance might be quite a bit more as MCBs do get a little warm in
operation.

On such short circuits what fault could you have that both gave
rise to a full short circuit and which did not implicate the cable.




--
Ed Sirett - Property maintainer and registered gas fitter.
The FAQ for uk.diy is at http://www.diyfaq.org.uk
Gas fitting FAQ http://www.makewrite.demon.co.uk/GasFitting.html
Sealed CH FAQ http://www.makewrite.demon.co.uk/SealedCH.html


  #12   Report Post  
Will Dean
 
Posts: n/a
Default

"Ed Sirett" wrote in message
news

In real life PSC of 6kA are uncommon although theoretically possible.
A plausible scenario would be a cleaners' socket in a meter cupboard on
the ground floor of a block of flats containing the sub-station in the
basement.


I quite agree, but the socket is an irrelevance really - I'm talking about
any S/C fault which occurs close to the start of a final circuit, and
therefore doesn't benefit from the reduction in fault level which the final
circuit gives.

I thought the resistance of the MCB might come to our rescue but I've just
measured one and it's DC resistance is less than 0.01 ohms as measured by
good test gear. Likewise the meter might help out a bit? However the AC
impedance might be quite a bit more as MCBs do get a little warm in
operation.


Pure reactance ("AC impedance") wouldn't create any heat.

On such short circuits what fault could you have that both gave
rise to a full short circuit and which did not implicate the cable.


Well, I'm really talking about cable faults in this case (e.g. someone bangs
something large and metallic right through the cable).

What I'm fishing for is the relevance/application of the adiabatic
calculations in such situations.

Will



  #13   Report Post  
Martin Angove
 
Posts: n/a
Default

In message ,
wrote:



I have been wary of MCBs in high PFC places ever since. Sub station was
through the wall to the CU. I suspect the PSC was in the region of 10 kA.



And as you will notice, most MCBs are rated to break 6kA, some (notably
the Wylex wire-replacement types) are lower.

To butt in a bit late, just to back up another poster, I've never
measured a PSC higher than about 1.6kA at the end of the meter tails,
and the majority are below 1kA. Possibly has something to do with not
living in the centre of London, but although it's an interesting
question, I'm not sure how relevant it is in "real life".

Taking the calculation that the first 2.3m or so of 2.5mm2 cable is "at
risk" and the fact that in order to create a short this close to a CU
the most likely method is a nail through the cable, is it really a
problem if the heat dissipation damages the cable? You're going to be
replacing the nail-damaged length anyway...

Hwyl!

M.

--
Martin Angove: http://www.tridwr.demon.co.uk/
Two free issues: http://www.livtech.co.uk/ Living With Technology
.... Get 'em by the balls and their hearts and minds will follow.
  #14   Report Post  
Andy Wade
 
Posts: n/a
Default

Will Dean wrote:

This is an abstract question - but I'm just interested to see what I've
misunderstood here...


Hi Will, sorry to come in a bit late to this. Hope you're still around...

Imagine a domestic installation, with a PSC of 6kA at the incoming supply,
fused at 100A. Right next to the incoming supply, there's a consumer unit,
which contains one 20A MCB. The CU is connected to the incoming supplies
with very short 25mmsq tails (I've ignored their contribution to lowering
the fault level.)

This 20A MCB supplies a short (let's say 0.5m) piece of 2.5mmsq T&E, which
supplies a single socket next to the consumer unit.

My calculations suggest that the L-N fault current at the socket is going to
be about 5kA, and for adiabatic I^2t or (kS)^2 compliance, we would need to
break a short circuit in around 3ms max to avoid damaging the 2.5mm cable.


All figures agreed.

This seems to be well off the bottom of graphs for MCB response times.

[...]

I'm clear why *load* is considered only at the end of a circuit, but not why
fault conditions are being calculated for the end of the circuit.

Given that fuse and MCB time vs. current graphs don't tend to go down to
single-millisecond levels, should one actually be looking at I2t let-through
graphs and comparing that with k2S2?


Yes, you raise some interesting points that a designer does need to be
aware of when the prospective fault level is high.

Firstly, the reason that the fault calculation is usually done at the
furthest point of the circuit is that that is /usually/ the worst case.
When we're not at the bottom of the graph, as you put it, a lower
fault current leads to higher I^2*t let-through and hence more risk of
cable damage.

There's a useful graphical approach he for any given cable CSA and
'k' value (k=115 for T&&E) you can superimpose an adiabatic withstand
line for the conductor on the operating characteristic plot for the fuse
or MCB. It's a log-log graph, so the cable's line is straight, with a
gradient of -2 (t proportional to 1/I^2). You then see at a glance the
minimum fault current at which the fuse or MCB will definitely protect
the cable, i.e. the point of intersection of the two lines. With some
(non-current-limiting) MCBs there may be a second intersection at high
fault level. At higher currents the device's curve is now above the
cable's line - meaning that the cable will not be protected.

So, yes, you do need to refer to manufacturer's I^2*t data, which will
almost certainly show that your bit of 2.5mm^2 wire is protected after
all. Most MCBs (and HRC fuses) now are 'current-limiting' - meaning
that at high current they will trip and quench the arc so quickly that
the instantaneous current never has time to build up to its full
prospective value. (Even if the fault occurs at the peak of the voltage
waveform the inevitable inductance in the circuit means that the current
has to rise from zero at a finite rate.)

Secondly, you might find that at 5kA the supplier's main fuse beats the
MCB - look at the characteristics for a 100A BS 1361 Type 2 fuse here
http://www.bussmann.co.uk/images/Dat...S1361/LR85.pdf and
extrapolate the line a little. Or use the I^2*t value given (57,300
A^2s) and conclude that this 100A fuse would protect a 2.5 mm^2 copper
conductor. This becomes an important point once the PSSC exceeds the
rated breaking capacity of the MCB - which is usually 6 kA or 9 kA (M6
or M9) - as the supplier's fuse is now acting as the backup protective
device.

Thirdly BS 7671 allows short lengths of conductor between a phase busbar
and the input side of a fuse/MCB to be excused from fault protection
provided that it's (a) 3 m long, (b) "erected in such a manner as to
reduce to a minimum the risk of fault current" and (c) "erected in such
a manner as to reduce to a minimum the risk of fire or danger to
persons". [Reg. 473-02-02]

OOI the highest PSSC I've encountered on a house installation is about
3.3 kA (Ze = 0.07 ohm). This is a property quite near the footway
(short service cable) with the substation a stone's throw down the road.


HTH
--
Andy
  #15   Report Post  
Will Dean
 
Posts: n/a
Default


"Andy Wade" wrote in message
...

Hi Will, sorry to come in a bit late to this. Hope you're still around...


Well, I've just come back...

Thanks for the intersting info, Andy.

At the time I made the original post, I only had a 15th edition to hand. I
have since bought a 16th (I have waited long enough in between for both my
15th and 16th to be brown-cover!), and it actually has explicit notes on the
MCB graphs about checking with mfg data for very short times.

As you rightly point out, the calculations for very small fractions of a
cycle *are* different in nature anyway.

Firstly, the reason that the fault calculation is usually done at the
furthest point of the circuit is that that is /usually/ the worst case.
When we're not at the bottom of the graph, as you put it, a lower fault
current leads to higher I^2*t let-through and hence more risk of cable
damage.


Hmm, I'm not sure I'm convinced about this by the kinds of calculations I've
done. Current is falling linearly with cable length, which brings I^2 down
very quickly. Because one almost always seems to be on the magnetic part of
the breaker curve, it's not clear to me that 't' is rising so quickly.

Again I suppose it's down to looking at real-world breaker characteristics.

OOI the highest PSSC I've encountered on a house installation is about 3.3
kA (Ze = 0.07 ohm). This is a property quite near the footway (short
service cable) with the substation a stone's throw down the road.


Yes, I'm aware that they tend in real life to be very low, but I wasn't
really concerned about the realities of domestic installations here - I was
merely trying to understand the proper way to do the calculations.

Cheers,

Will





  #16   Report Post  
Andy Wade
 
Posts: n/a
Default

Will Dean wrote:

At the time I made the original post, I only had a 15th edition to hand. I
have since bought a 16th (I have waited long enough in between for both my
15th and 16th to be brown-cover!), and it actually has explicit notes on the
MCB graphs about checking with mfg data for very short times.


Indeed so.

As you rightly point out, the calculations for very small fractions of a
cycle *are* different in nature anyway.


Firstly, the reason that the fault calculation is usually done at the
furthest point of the circuit is that that is /usually/ the worst case.
When we're not at the bottom of the graph, as you put it, a lower fault
current leads to higher I^2*t let-through and hence more risk of cable
damage.


Hmm, I'm not sure I'm convinced about this by the kinds of calculations I've
done. Current is falling linearly with cable length, which brings I^2 down
very quickly. Because one almost always seems to be on the magnetic part of
the breaker curve, it's not clear to me that 't' is rising so quickly.


You're right. For MCBs I^2*t increases with /falling/ current below the
'instantaneous' trip point (i.e. on the thermal-trip part of the
characteristic) but increases with /rising/ current on the magnetic-trip
part of the characteristic. There's a graph of I^2*t against I on page
130 of the /commentary/.

[...] I was merely trying to understand the proper way to do the
calculations.


Me too... You would find the commentary interesting (well I did):
http://www.iee.org/Publish/Books/Wir...?book=NS%20031

--
Andy
  #17   Report Post  
Will Dean
 
Posts: n/a
Default

"Andy Wade" wrote in message
...

You're right. For MCBs I^2*t increases with /falling/ current below the
'instantaneous' trip point (i.e. on the thermal-trip part of the
characteristic) but increases with /rising/ current on the magnetic-trip
part of the characteristic. There's a graph of I^2*t against I on page
130 of the /commentary/.


Interesting. That suggests that one should be very careful about where one
does the adiabatic calculations - it's certainly a valuable footnote when
demonstrating compliance by applying that calculation only at the far end of
the circuit, if you've assumed that the magnetic + limited-let-through saves
you at the near end.

Me too... You would find the commentary interesting (well I did):
http://www.iee.org/Publish/Books/Wir...?book=NS%20031


Grrrr. As I've just updated both my wiring regs and OSG, and bought BS7909
( GBP4.30 per page!) in the last couple of weeks, my standards budget is
rather blown at the moment. I'd like GN3, as well.

But it's useful to have a recommendation for the book - some of the 'helper'
books that go with the regs are poor, IMO. Paul Cook seems pretty good, in
general, though. And I do get an IEE discount.

I'll just about get the set up to date, and they'll issue a '17th edition',
no doubt.

Cheers,

Will




  #18   Report Post  
Andy Wade
 
Posts: n/a
Default

Will Dean wrote:

But it's useful to have a recommendation for the book - some of the 'helper'
books that go with the regs are poor, IMO.


Err, yes. They could combine all the GNs into one book for a start.
Also we now have the OSG (essential but poorly constructed) /and/ the
Electrician's Guide to the building Regs. There's a huge amount of
overlap between these two and they cry out to be combined.

Paul Cook seems pretty good, in general, though.


Yes, although Brian Jenkins, who did the commentary on the 15th Edition,
is a better writer, IMHO.

I'll just about get the set up to date, and they'll issue a '17th edition',
no doubt.


There may not ever be a 17th Ed. - just further editions of BS 7671, I
guess. Then BS EN 6xxxx and maybe BS EN ISO xxxxxx. Who knows?

--
Andy
Reply
Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules

Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
how can a light bulb become a short circuit? dstromb Electronics Repair 5 May 1st 05 11:03 PM
Ink Jet Driver Circuits Blown Due to Short Circuit Caused By Ink Spill? Dieter Schmitz Electronics Repair 0 December 19th 04 08:20 PM
Circuit breaker did not t barry martin Home Repair 1 September 20th 04 04:32 AM
Strange not-quite-240 high-voltage smoke problem... short? circuit breaker? Kent Monroe Home Repair 10 February 22nd 04 02:42 AM
Audio amp circuit help? Ken Taylor Electronics Repair 3 December 3rd 03 02:04 PM


All times are GMT +1. The time now is 03:44 AM.

Powered by vBulletin® Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
Copyright 2004-2024 DIYbanter.
The comments are property of their posters.
 

About Us

"It's about DIY & home improvement"